Example 2 
Let \(B\)  be a left cancellative left semi-braces of size \(pq\)  (where \(p,q\)  are prime numbers, \(p>q\) ). Then, if \((B,\circ )\)  is abelian or \(|E|=p\) , since \(G\)  is a trivial skew left brace of size \(q\) , by [1]} we have that \(B\)  is left nilpotent. If \((B,\circ )\)  is not abelian and \(|E|=q\) , then \(B\)  is a semidirect product of left semi-braces as in teostrutt. By a standard calculation, we obtain that \(B^2=B\) , hence \(B\)  is not left nilpotent.

